Understanding the Problem
To determine the mass of nitrogen ($N_2$) required to produce a specific amount of ammonia ($NH_3$), we use the principles of stoichiometry. Stoichiometry allows us to relate the amounts of reactants and products in a chemical reaction based on their balanced chemical equation.
The balanced equation provided is:
$N_2 + 3H_2 \rightarrow 2NH_3$
This equation tells us that 1 mole of nitrogen gas reacts to produce 2 moles of ammonia.
Step-by-Step Solution
1. Calculate the number of moles of ammonia ($NH_3$)
We are given $6.022 \times 10^{24}$ molecules of $NH_3$. We use Avogadro's number ($N_A = 6.022 \times 10^{23}$ molecules/mol) to convert this into moles:
$n(NH_3) = \frac{\text{Number of molecules}}{N_A} = \frac{6.022 \times 10^{24}}{6.022 \times 10^{23}} = 10 \text{ moles}$
2. Determine moles of Nitrogen ($N_2$) required
From the balanced equation, 2 moles of $NH_3$ require 1 mole of $N_2$. Therefore, 10 moles of $NH_3$ require:
$n(N_2) = \frac{1}{2} \times n(NH_3) = \frac{1}{2} \times 10 = 5 \text{ moles of } N_2$
3. Calculate the mass of Nitrogen ($N_2$)
The molar mass of $N_2$ is approximately $28.0 ext{ g/mol}$ (since atomic mass of N is $14.0 ext{ g/mol}$).
$\text{Mass} = n \times \text{Molar Mass}$ $\text{Mass} = 5 \text{ mol} \times 28.0 \text{ g/mol} = 140 \text{ grams}$
Conclusion
To produce $6.022 \times 10^{24}$ molecules of ammonia, you require 140 grams of nitrogen gas.